91Ó°ÊÓ

The given data can be listed below as:

• The mass of the truck engine is$\text{m}=409\text{kg}$.
• The angle made by cable C is$\mathrm{θ}=37.1°$.
• The tension in cable A is${\text{T}}_{\text{A}}=722\text{N}$.

The tension is described as the pulling force that is transmitted in an axial direction with the help of a string or pulley. It is also referred to as the pair of the action and the reaction force.

The free body diagram of the system has been drawn belo

Here, ${\text{T}}_{\text{A}}$is the tension of cable A, ${\text{T}}_{\text{B}}$ is the tension of cable B, and ${\text{T}}_{\text{C}}$is the tension of cable C.

All the forces in the and in the direction is zero.

The equation of the forces acting in the x-direction is expressed as:

$$\begin{gathered} {\text{T}}_{\text{A}}={\text{T}}_{\text{C}}\text{²õ¾±²Ôθ} \\ {\text{T}}_{\text{C}}=\frac{{\text{T}}_{\text{A}}}{\text{²õ¾±²Ôθ}} \end{gathered}$$

Here, ${\text{T}}_{\text{A}}$is the tension of cable A, ${\text{T}}_{\text{C}}$is the tension of cable C, and $\mathrm{θ}$is the angle made by cable C.

Substitute the values in the above equation.

$\begin{array}{rcl}{\text{T}}_{\text{C}}& =& \frac{722\text{N}}{\sin 37.1°}\\ & =& \frac{722\text{N}}{0.603}\\ & =& 1197\text{N}\\ & & \end{array}$

The equation of the forces acting in the y-direction is expressed as:

${\text{T}}_{\text{B}}={\text{T}}_{\text{C}}\text{³¦´Ç²õθ}+{\text{T}}_{\text{D}}$

Here, ${\text{T}}_{\text{D}}$is the tension of cable D.

As the weight of the engine is the tension of the cable D, then the above equation can be expressed as:

${\text{T}}_{\text{B}}={\text{T}}_{\text{C}}\text{³¦´Ç²õθ}+\text{mg}$

Here, $\mathrm{m}$is the mass of the truck engine and $\mathrm{g}$is the acceleration due to gravity.

Substitute the values in the above equation.

$\begin{array}{rcl}{\text{T}}_{\text{B}}& =& \left(1197\text{N}\right)\cos 37.1°+\left(409\text{kg}\right)\left(9.8{\text{m/s}}^{\text{2}}\right)\\ & =& \left(1197\text{N}\right)\left(0.797\right)+4008.2\text{kg}·{\text{m/s}}^{\text{2}}\\ & =& \left(954.009\text{N}\right)+4008.2\text{kg}·{\text{m/s}}^{\text{2}}×\frac{1\text{N}}{1\text{kg}·{\text{m/s}}^{\text{2}}}\\ & =& 4963\text{N}\\ & & \end{array}$

Thus, the tensions in cables B and C are $4963\text{N}$and $1197\text{N}$respectively.